PERL：我从本地时间获取 mday 值，如下所示。我如何从本地时间获取的 mday 中减去 1

Question

PERL：我从 locatime 中获取 mday 值如下，现在我想要前天的值。我如何从本地时间

中获取的 mday 中减去 1

my ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime();
my $part = "P".$mday;

print "Today value is $part \n";

my $part_yes = "P".$mday - $num;

print "$part_yes \n";

Answer 1

my $now = time(); # time as seconds since 1970-01-01
my $day_ago = $now - 24*60*60;

my ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime($day_ago);
my $part = "P".$mday;

警告（ikegami 评论）：并非所有的日子都有 24 小时。这可以产生提前 0、1 或 2 天的结果

https://perldoc.perl.org/functions/time
https://perldoc.perl.org/functions/localtime

Answer 2

使用日期时间：

my $dt =
   DateTime
   ->now( time_zone => 'local' )
   ->set_time_zone('floating')  # Do this when working with dates.
   ->truncate( to => 'days' );  # Optional.

$dt->subtract( days => 2 );

my $yesterday = $dt->day;

DateTime 非常繁重，似乎人们问 date-time 问题总是回来说“只有核心模块！”，所以这里有一个只使用核心模块的解决方案。

use Time::Local qw( timegm );

# Create an timestamp with the same date in UTC as the one local one.
my $epoch = timegm(0, 0, 0, ( localtime() )[3,4,5]);

# We can now do date arithmetic without having to worry about DST switches.
$epoch -= 2 * 24*60*60;

my $yesterday = ( gmtime($epoch) )[3] + 1;

PERL：我从本地时间获取 mday 值，如下所示。我如何从本地时间获取的 mday 中减去 1

PERL : I take mday value from locatime as below , . How can I subtract 1 from mday which I take from localtime

perl

date

time

timezone

nls